abstract algebra - Proving that either $x$ or $2x$ is a generator of the cyclic group $(mathbb Z_3[x]/langle f(x) rangle)^*$

if $f(x)$ is a cubic irreducible polynomial over $\mathbb Z_3$, prove that either $x$ or $2x$ is a generator of the cyclic group $(\mathbb Z_3[x]/\langle f(x) \rangle)^*$

Attempt: $f(x) = \alpha x^3+ \beta x^2 + \gamma x + \rho~~|~~ \alpha ,\beta,\gamma,\rho \in \mathbb Z_3$

$\implies \mathbb Z_3[x]/\langle f(x) \rangle = \{ax^2+bx+c~~|~~a,b,c \in \mathbb Z_3\}$

$O[ \mathbb Z_3[x]/\langle f(x) \rangle ]^*=26$

Hence, elements can $(\mathbb Z_3[x]/\langle f(x) \rangle)^*$ can have orders $1,2,13,26$.

If $x$ is the generator for the above field, then $x^{26}=1$, then : $(2x)^{26} = (-x)^{26}=1$. Hence, isn't $2x$ also a generator, contrary to what the question is asking?

How do I proceed ahead?

Thank you for your help.

Answer

I'll denote the image of $x$ in the quotient as $\bar{x}$.

It's clear that $\bar{x}$ doesn't have order 1, and similarly it doesn't have order 2 (because then $\bar x$ would be a root of $x^2 - 1 \in \Bbb{Z}_3[x]$, contradicting that the cubic $f(x)$ (which has $\bar x$ as a root) was irreducible).

Thus $\bar x$ has order 13 or 26.

Likewise, $2\bar x$ doesn't have order 2 as this would imply $\bar x$ had order 2. So $2\bar x$ also has order 13 or 26.

If $\bar x$ has order 26 we're done, so suppose $\bar x$ has order 13, so $\bar x^{13} = 1$. Then $(2\bar x)^{13} = 2\bar x^{13} = 2$, so $2\bar x$ must have order 26, and generates the group.